/sci/ - Science & Math » Thread #13301213

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Anonymous Sun 20 Jun 16:07:21 2021 No.13301213 View Reply Original Report

Quoted By: >>13302606 >>13302614 >>13302650 >>13302670 >>13302899

Say I wanted to run an iterative function of ?x in order to produce n super roots of 2. What is simple code in order to calculate this if I want to see if recursively doing this would eventually result in an integer? Doing this by calculator converged towards 1 with an increasing number of 0's after the decimal point, so somewhere in there an integer might be producible.

Anonymous

Anonymous Sun 20 Jun 2021 21:57:21 No.13302606 Report

Quoted By:

>>13301213
No idea, lmao.
But BUMP so you get an answer.

Anonymous

Anonymous Sun 20 Jun 2021 21:59:01 No.13302613 Report

Quoted By: >>13302670 >>13302832

do you mean sqrt(sqrt(sqrt(...(x))))?

Anonymous

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Anonymous Sun 20 Jun 2021 21:59:14 No.13302614 Report

Quoted By: >>13302832

>>13301213
>n super roots of 2
What is a super root?

Anonymous

Anonymous Sun 20 Jun 2021 22:07:22 No.13302650 Report

Quoted By: >>13302845

>>13301213
computers cant give you infinite precision so you cant evaluate sqrt(2) accurately. That implies that you wont be able to find this using a computer. you can probably show that the root of an irrational number is irrational analytically

Anonymous

Anonymous Sun 20 Jun 2021 22:11:07 No.13302670 Report

Quoted By:

>>13301213
If you just mean >>13302613 then it's trivial to show that $x^{({1 / 2})^n}$ tends to 1 for very big n, since the geometric series with common ratio <1 tends to 0 and $x^0$ is 1
But you'd never get an integer at any level of iteration beyond the first few

Anonymous

Anonymous Sun 20 Jun 2021 23:01:58 No.13302832 Report

Quoted By: >>13302845

>>13302613
Yes, this exactly.
>>13302614
It's the reverse of tetration(hyper4) or pentration(hyper5) which is exponents to themselves 4 or 5 times. Represented like n^n^n^n^n^... and so on.

Anonymous

Anonymous Sun 20 Jun 2021 23:07:50 No.13302845 Report

Quoted By:

>>13302832
first see >>13302650
you can use recursive bijections to approximate a square root to whatever accuracy you want
then you can just do it again.

Anonymous

Anonymous Sun 20 Jun 2021 23:28:14 No.13302899 Report

Quoted By: >>13302912

>>13301213
If it’s converging from 2 to 1, how could any integer be produced interstitially?

Anonymous

Anonymous Sun 20 Jun 2021 23:33:25 No.13302912 Report

Quoted By: >>13302953

>>13302899
According to the following its unknown.
https://en.wikipedia.org/wiki/Tetration
>Open questions
>Other than the problems with the extensions of tetration, there are several open questions concerning tetration, particularly when concerning the relations between number systems such as integers and irrational numbers:
>It is not known whether there is a positive integer n for which n? or ne is an integer. In particular, it is not known whether either of 4? or 5e is an integer.[citation needed]
>It is not known whether nq is an integer for any positive integer n and positive non-integer rational q.[21] For example, it is not known whether the positive root of the equation 4x = 2 is a rational number.[citation needed]

Anonymous

Anonymous Sun 20 Jun 2021 23:46:18 No.13302953 Report

Quoted By:

>>13302912
that’s not what that says.
1 < ?2 < 2
n > 1 -> ?n > 1
no integer between 1 and 2

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